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Diketahui larutan asam benzoat  0,025M mempunyai p...

Diketahui larutan asam benzoat begin mathsize 14px style open parentheses C subscript 6 H subscript 5 C O O H close parentheses end style 0,025M mempunyai pH yang sama dengan larutan begin mathsize 14px style H subscript 2 S O subscript 4 end style 0,0005M. Tentukan harga begin mathsize 14px style K subscript a end style asam benzoat!

Jawaban:

Asam benzoat begin mathsize 14px style open parentheses C subscript 6 H subscript 5 C O O H close parentheses end style termasuk asam lemah, sedangkan begin mathsize 14px style H subscript 2 S O subscript 4 end style termasuk asam kuat. Karena kedua asam tersebut memiliki pH yang sama, maka undefined sama.

Rumus  undefined pada begin mathsize 14px style H subscript 2 S O subscript 4 end style :

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals M cross times a space space space space space space space space equals 5 cross times 10 to the power of negative sign 4 end exponent cross times 2 space space space space space space space space equals 10 to the power of negative sign 3 end exponent end style

Rumus  undefined pada begin mathsize 14px style C subscript 6 H subscript 5 C O O H end style :

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript a cross times M end root end cell row cell 10 to the power of negative sign 3 end exponent end cell equals cell square root of K subscript a cross times 25 middle dot 10 to the power of negative sign 3 end exponent end root end cell row cell 10 to the power of negative sign 6 end exponent end cell equals cell K subscript a cross times 25 middle dot 10 to the power of negative sign 3 end exponent space space space space end cell row cell K subscript a end cell equals cell fraction numerator 1 middle dot 10 to the power of negative sign 6 end exponent over denominator 25 middle dot 10 to the power of negative sign 3 end exponent end fraction end cell row cell K subscript a end cell equals cell fraction numerator 100 middle dot 10 to the power of negative sign 8 end exponent over denominator 25 middle dot 10 to the power of negative sign 3 end exponent end fraction end cell row cell K subscript a end cell equals cell 4 middle dot 10 to the power of negative sign 5 end exponent end cell end table end style  

 

Jadi, begin mathsize 14px style K subscript a end style asam benzoat adalah begin mathsize 14px style 4 middle dot 10 to the power of negative sign 5 end exponent end style

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